as key leads to the entry being garbage collected.Huh!Tried all combinations. No clarity. Last resort - Pinged Rajiv.And so went the conversation -

Me : If you put ("abc" + "def").intern(); as key, it doesnt get GC'd but if you put (new String("abc") + "def").intern() it gets GC'dRajiv : Decompile and see if "abc"+"def" is being converted to "abcdef" by javacMe: Yes it is. So?[This could be the clue. Am still thinking.. tick tick tick]

Rajiv: Check if "abcdef"== (new String("abc") + "def").intern()Me: It is... printed the identitity hashcodes.Rajiv: In the class you have both "abcdef" and (new String("abc") + "def").intern() and still (new String("abc") + "def").intern() gets gc'ed?Me: God! Then it doesn't get gc'd.[Now Rajiv cracks it -]

Rajiv:"I think intern is weak map and constant pool has a strong ref"Me: ohh!Me: In that case (new String("abc") ).intern(); should get GC'd right? But we saw it doesn't. The maya happens only when someString is '+'d to (new String("abc")) and then the resultant String is interned.Me: Just (new String("abc")).intern() doesnt get GC'd.Rajiv: When you say (new String("abc")).intern() there is a string "abc" in constant pool.Me: Yes "abc" in constant pool would be the literal we created and passed as argument to the String constructor.Rajiv: (new String("abc")).intern() returns that string. So wont get gc'edMe: Oh yeah. Got it!Me: So only when you do a "+" you get a String which is not there in constant pool and hence it gets GC'd ...Rajiv: ya right.

I had earlier thought of intern pool and constant pool to be the same. But Rajiv 's prediction of intern being a weak map and constant pool holding a strong ref looks quite convincing.Oo la.. That solved our mystery. Thanks Rajiv:-)

while (true) { System.gc(); /** * Verify Full GC with the -verbose:gc option * We expect the map to be emptied as the strong references to * all the keys are discarded. */ System.out.println("map.size(); = " + map.size() + " " + map); } }}

What do we expect the size of the map to be after full GC? I initially thought it should be empty. But it turned out to be 2.

Look at the way the four Strings are initialized. Two of them are defined using the 'new'operator, whereas the other two are defined as literals. The Strings defined using the 'new' operator would be allocated in the Java heap, but the Strings defined defined as literals would be in the literal pool.The Strings allocated in the literal pool (Perm Space) would never be garbage collected.This would mean that String 'str2' and 'str3' would always be strongly referenced and the corresponding entry would never be removed from the WeakHashMap.

So next time you create a 'new String()', put it as a key in a WeakHashMap, and later intern() the String, beware - Your key will always be strongly referenced. [Invoking intern() method on a String will add your String to the literal pool if some other String equal to this String does not exist in the pool]

//Discard the strong reference to the key name = null; while (true) { System.gc(); /** * Verify Full GC with the -verbose:gc option Since there is no strong reference to the key, it is assumed that the entry has been removed from the WeakHashMap */ System.out.println(cache.size()); } }

Now when the testMethod() is run what do you expect the output to be? Since the strong reference to key is discarded, we assume that the entry from the map would be removed, and map would be empty after a full GC.But that does not happen though.

Let us see what was the put operation on the WeakHashMap.

cache.put(name, new ComplexDO("1", name));

Here the value ComplexDO was holding the key name. This would mean that the value always strongly refers to the key, and hence the key would never be garbage collected. The entry would always remain the map.

This is what WeakHashMap API says - "The value objects in a WeakHashMap are held by ordinary strong references. Thus care should be taken to ensure that value objects do not strongly refer to their own keys, either directly or indirectly, since that will prevent the keys from being discarded."

//Since s1.equals(s2) is true and hash is same, the earlier value //against key s1 ("good") in the map is replaced by the new one. ("ok")

8 s1=null;

9 System.gc(); //Verify Full GC with the -verbose:gc option

10 System.out.println(map.size());11 }12 }

What do we expect the output to be? 1? No, Not exactly.

Here s1 and s2 are two different objects on the heap. So in line 5, a new (key,value) pair with key s1 is put into the map. Later when a (key,value) with key s2 is being put into the map, it checks for equals on s1 and s2 and their hashcode. When it finds the equals returns true and hashcde is same, it replaces the value of the earlier entry with the new value. But the issue(?) here is, WeakHashMap/HashMap does not replace the earlier key while adding a (key, value) pair whose key is actually a duplicate key in the map.So even after putting an entry with key s2, the WeakHashMap has only one entry whose key refers to the object refered by s1 and not s2.Now the object on the heap refered by s1, has one strong reference(through s1) and one weak reference through the WeakHashMap.Later when I say s1=null, the object on the heap refered to by s1 lost the strong reference and when gc happens, the entry is removed from the map.

So thats how it works.

Also note WeakHashMap is only a wrapper over HashMap and the HashMap's put api says " If the map previously contained a mapping for this key, the old value is replaced by the specified value."

So just be careful when you use WeakHashMap and your usage scenario is similar to the above.